The solution of the differential equation x y | vedclass

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(B) The given differential equation is $x y^2 d y = (x^3 + y^3) d x$.This can be rewritten as $\frac{d y}{d x} = \frac{x^3 + y^3}{x y^2}$.This is a ho

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$y^3 = 3 x^3 \log (c x)$

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(B) The given differential equation is $x y^2 d y = (x^3 + y^3) d x$.This can be rewritten as $\frac{d y}{d x} = \frac{x^3 + y^3}{x y^2}$.This is a homogeneous differential equation. Let $y = v x$, then $\frac{d y}{d x} = v + x \frac{d v}{d x}$.Substituting these into the equation:$v + x \frac{d v}{d x} = \frac{x^3 + v^3 x^3}{x(v x)^2} = \frac{x^3(1 + v^3)}{x^3 v^2} = \frac{1 + v^3}{v^2}$.$x \frac{d v}{d x} = \frac{1 + v^3}{v^2} - v = \frac{1 + v^3 - v^3}{v^2} = \frac{1}{v^2}$.Separating the variables, we get $v^2 d v = \frac{1}{x} d x$.Integrating both sides: $\int v^2 d v = \int \frac{1}{x} d x$.$\frac{v^3}{3} = \log |x| + C$, where $C = \log c$.$\frac{v^3}{3} = \log |x| + \log c = \log |c x|$.Substituting $v = \frac{y}{x}$, we get $\frac{1}{3} (\frac{y}{x})^3 = \log |c x|$.$\frac{y^3}{3 x^3} = \log |c x| \Rightarrow y^3 = 3 x^3 \log |c x|$.

The solution of the differential equation $x y^2 d y - (x^3 + y^3) d x = 0$ is

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